Find the equivalent impedance Zeq seen by the source when Vs = 5 cos (5t) v, C = 2 F, R = 3 Ω and L = 0.5 H. (Give angles in degrees and round final answers to two decimal places.) Calculate the voltage across the capacitor. (Give angles in degrees and round final answers to two decimal places.) complex impedance The equivalent impedance seen by the source is 3 + j 2.4 Ω = 3.84 at an angle of 38.66 Ω. The voltage across the capacitor is at an angle of -128.66 v.

The circuit impedance is 3.84 phase 38.65º and the voltage across the capacitor is 0.13 phase -128.65º V.

Explanation:

Since the voltage given to us was Vs = 5*cos(5t) V and it is the form of V = Vmax*cos(omega*t) V we can extract the frequency omega, wich is w = 5 rad/s.

In the circuit we have a capacitor and a inductor. The capacitor impedance is negative and it is inversely proportional to the frequency, while the inductor impedance is positive and directly proportional to the frequency. So we have:

Z = R + jw*L - j/(wC)

Z = 3 + j*5*0.5 - j/(5*2)

Z = 3 + j*2.5 - j*0.1 = 3 + j*2.4 Ohm = 3.84 phase 38.65º Ohm

To find out the voltage across the capacitor we can use a voltage divider equation that is:

Vcapacitor = [Zcapacitor/(R + Zinductor + Zcapacitor)] * Vsource

Vcapacitor = [(-j0.1)/(3 + j2.4)]*Vsource

Vcapacitor = [(0.1 phase -90º)/(3.84 phase 38.65º)]*5 phase 0º

Vcapacitor = [0.026 phase -128.65º]* 5 phase 0º

Vcapacitor = 0.13 phase -128.65º V

Given Information:  

Source voltage = Vs = 5cos(5t) V

Resistance = R = 3 Ω

Capacitance = C = 2 Farads  

Inductance = L = 0.5 H  

Frequency = ω = 5 rad/sec  

Required Information:  

Equivalent impedance = Zeq = ?

Voltage accross capacitor = Vcap = ?

Equivalent impedance = Zeq = 3.84 < 38.66° Ω

Voltage accross capacitor = Vcap = 0.13 < -128.66° V

Explanation:

The source voltage = 5cos(5t)

In polar form,

Vs = 5 < 0° V

(a) The equivalent impedance is the sum of resistance, capacitive reactance, and inductive reactance

Zeq = R + Xc + XL

Where Capacitive reactance is given by

Xc = 1/jω*C

Xc = 1/j*5*2

Xc = -j0.1 Ω  

in polar form,

Xc = 0.1 < -90° Ω

The inductive reactance is given by

XL = jω*L  

XL = j*5*0.5

XL = j2.5  Ω

in polar form,

XL = 2.5 < 90° Ω

Therefore, the equivalent impedance can now be found

Zeq = R + Xc + XL

Zeq = 3 - j0.1 + j2.5

Zeq = 3 + j2.4 Ω  

In polar form,

Zeq = 3.84 < 38.66° Ω

(b) The voltage across the capacitor can be found by using the voltage divider rule

Vcap = Vs [ Xc/Zeq]

Where Vs is the source voltage, Xc is the impedance of capacitor, and Zeq is the impedance of overall circuit.

Vcap = (5<0) [0.1<-90°/3.84<38.66°]

Vcap = 0.13 < -128.66° V